CHAPTER 19 CURRENT, RESISTANCE AND ELECTROMOTIVE FORCEpeople.physics.tamu.edu/adair/phys202/CHAPTER...

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1 CHAPTER 19 CURRENT, RESISTANCE AND ELECTROMOTIVE FORCE BASIC CONCEPTS CURRENT and CURRENT DENSITY RESISTANCE and RESISTIVITY BATTERY INTERNAL RESISTANCE ENERGY AND POWER IN CIRCUITS

Transcript of CHAPTER 19 CURRENT, RESISTANCE AND ELECTROMOTIVE FORCEpeople.physics.tamu.edu/adair/phys202/CHAPTER...

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CHAPTER 19

CURRENT, RESISTANCE AND

ELECTROMOTIVE FORCE

BASIC CONCEPTS

CURRENT and CURRENT DENSITY

RESISTANCE and RESISTIVITY

BATTERY INTERNAL RESISTANCE

ENERGY AND POWER IN CIRCUITS

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CURRENT IS THE MOVEMENT OF CHARGE

IN A MATERIAL

In some cases the objects that are moving

are positive.

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In other cases, for example in metals, the

objects are negative (electrons).

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Current is the time rate of passage of the

charge.

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RESISTIVITY

Resistivity, ρ, is a characteristic of a

material.

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RESISTIVITY

Resistivity, ρ, changes with temperature.

���� = ���1 + �� − ����

Where �� is the value at 200 C.

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For a metal:

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For Semiconductor:

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Some materials loose all resistivity –

Superconductors:

Discovered in 1911 by H. Kammerlingh

Onnes.

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RESISTANCE

Resistance is a characteristic of an object.

Resistance, R, is related to Resistivity, ρ, by

= ��

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The longer the object the higher the

resistance.

The larger the cross-section the smaller the

resistance.

If ρ is constant the total current through a

conductor is proportional to the voltage

across it.

� ∝ �

Or

� ∝ �

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The constant of proportionality is the

resistance, R.

� = �

Or

� = �

This is Ohm’s Law.

If a resistor obeys Ohm’s Law it is an Ohmic

resistor.

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Obeys Ohm’s Law:

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Semiconductors are not Ohmic:

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POWER DISIPATED WITH CURRENT FLOW.

Remember

q

V

Change in energy ∆�

∆� = ��

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� = �

Or � =�

� = ��

� = ��

� =��

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CALCULATE RESISTANCE

A copper rod with cross section A has a

length L what is its resistance? Work for

� = 3.14�10� !� and � = 10!.

From Table 25.1 � = 1.72�10�$Ω!

Therefore

= ��

�= �1.72�10�$Ω!�

10!

3.14�10� !�

= 5.48�10��Ω

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The figure is of a 12 V battery with no

current in the circuit.

The voltmeter will read 12 V.

But if there is current:

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The voltmeter will read less than 12 V.

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What is the current in the circuit?

� = �

� =�

=

12

4 + 2= 2�

What will the voltmeter read?

�()*+ = 12� − � ,-.)(-*/

�()*+ = 12� − �2���2Ω�

�()*+ = 12� − 4� = 8�

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DIRECT-CURRENT CIRCUITS

BASIC CONCEPTS

Resistors in circuits.

Kirchhoff’s Rules

Ammeters and Voltmeters

R-C Circuits

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In Chapter 24 we found equivalent

capacitance for combinations of capacitors.

1

0)1=

1

02+

1

0�456786987

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0)1 = 02 + 0�456:;6;<<8<

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For resistors we need similar equations.

The current I through each R is same.

The voltage �*= = �*> + �>? + �?=

And � = �

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Therefore

� 2 + � � + � @ = � )1A,B*/)-.

Can replace the three resistors with one

resistor with resistance )1A,B*/)-.

)1A,B*/)-. = 2 + � + @456786987

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The total current I is divided between

2, �, ;DE @

� = �2 + �� + �@

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And

� =�

Therefore

�FGHIJKLFMN=

�O+

�P+

�Q

Or

1

)1A,B*/)-.=

1

2+

1

�+

1

@

For Parallel.

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Consider the following circuit.

What is the current through the 6 Ω

resistor?

12V

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��R =��R

2Ω=12�

2Ω= 6�

Therefore there are 6A through the 1Ω

resistor.

�2T = �2R1Ω = 6�1Ω = 6�

The emf of the battery is

U = �2T + ��R = 6� + 12� = 18�

The current through the 6Ωresistor is

� R =U

6Ω=18�

6Ω= 3�

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Example:

1.25A

What does the voltmeter read?

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KIRCHHOFF’S RULES

Junction Rule

The sum of the currents into a junction is

zero.

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Loop Rule

The sum of the potential differences in a

loop is zero.

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Sign Convention

For resistors:

For batteries:

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Example:

Loop 1 (start at a)

−�� + U2 − �262 = 0

Loop 2 (start at c)

+U2 − �262 − ��6� − U� = 0

I1

I2

IR

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Loop 3 (start at a)

−�� + U� + ��6� = 0

Junction a

�2 = �� + ��

Junction b

�� + �� = �2